 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2, module 4. So, in past three modules we have been preparing for proof of implicit and inverse function theorem. Last time we also saw the statement of the implicit function theorem. So, let me just recall the statement again. I have explained the statement last time. So, let me just recall this one now only. v and w are Banach spaces, y is any topological space, m cross n be an open subset of y cross v and a function capital F from m cross n to w which is continuous. Then it satisfies three more conditions. What are they? For some point y naught v naught inside m cross n, f of y naught v naught is 0. For each y inside m, the function f y which stands v to capital F y v that is differentiable as a function from n to w. And this derivative for each y I have, so the derivative function from m cross n to v v w that is continuous. And the derivative at y naught v naught namely g y naught v naught is similarity between v and w. In particular v and w are similar Banach spaces, isometric Banach spaces. Then there exists rho positive and an open nibbled m prime of y naught inside m and a function g from the closed wall of radius rho around v naught. Oh sorry, from m prime to the closed wall such that f of y g y is 0. Moreover this function g from m prime to b rho b rho bar v naught is continuous. This is the first conclusion. The second conclusion, conclusion b here requires one more hypothesis namely assume y is also a Banach space and the function restricted to y naught v naught from this m to w namely f y v naught operating upon y is just capital F of y v naught. This is differentiable at y naught and its derivative is h. Then g will be differentiable at y naught and the derivative of g is given by minus t inverse h. So, let us start proving this one now. There are a number of steps to be taken so that we understand what is going on. So, the proof is broken up into smaller steps. The first step is I want to make a simplification in the statement as well as in the proof namely I would like to bring this t to be identity map. How can I do that? Namely by composing with t inverse from v w and as if we are working now all the time inside v ok. Compose with t inverse do not go to v at all w at all keep coming into v. That means you can assume that v and w are actually the same space instead of similarity. So, how do how this is what I want to do? So, how do I do that? So, as follows namely put f hat replace f by f hat which is t inverse of f. Remember f was from m cross v m cross n to w now it will be n cross n to v itself. So, that is all. Now, suppose we have proved this theorem to f hat in this special case then we can go back to the original by pre again composing by t. So, if we replace f by f hat then the derivative of this one will be now t t inverse into t which is the identity at the point y naught v naught ok. Because the derivative of a linear automorphism whatever it is itself when you by composition law by by chain rule derivative of this f t inverse composite f will be t inverse composite derivative of f. So, that will be identity. So, this is what we want to assume so that writing down the proof will become easier. I do not have to keep on writing t here that is all ok. So, that is the first step all right. Maybe we will do use this one only for a as soon as we have proved it for this special case we know that it is true for the general case also because all that I have to do is apply the same thing t composite that thing to get back to over f ok. In the step two now we have the modified hypothesis all 1, 2, 3 are all modified namely this now w is v itself. So, m cross v to v we have got a function instead of m cross v to w ok. So, I am writing m cross v to v now you take the function new function s defined by s of y v equal to v minus f of y v f is a function from m cross n to w ok. So, define s of y v equal to v minus f of y v ok. For every point the f is defined in the army where f was defined from m cross n. So, this function will be defined on now m cross n again ok. For every point 0 less than for every epsilon between 0 and 1 let us have this short notation instead of writing all the times minimum of epsilon and half ok c epsilon. We claim that there exist an open set m prime such that y naught is inside m prime contained inside m and the positive row such that s restricted to the restricts to a function from m prime cross the closed ball goes inside the closed ball and satisfies this inequality this is our first step. Part of this remember is existence of this m prime and row this was a part of a right. But the conclusion is slightly you know in between conclusion is a weaker conclusion we are not yet saying that there is a unique g and etcetera g has not yet appeared. So, first thing is the new function s has this property named s of y v 1 minus s of y v 2 is less than c epsilon times v 1 minus v 2 for every y m prime v 1 and v 2 are inside the closed ball. Remember this was nothing but a uniform contraction. So, so we are up to now apply contraction mapping which was done in the first module remember that right. So, first we have to claim this one. So, first let us get the proof of this part ok. So, that is the ok. So, how to prove this one? g of y naught v naught is identity now right earlier it was just a similarity t in the new hypothesis is identity. By continuity of g we first select a neighborhood m prime of y and a row positive such that g y v minus identity is less than c epsilon. So, this is where the continuity of the derivative in the second variable that is used ok. This g was the differentiation of capital F right with respect to the second variable v. So, by continuity of this some neighborhood of y naught and some neighborhood of v naught will go inside that. In the neighborhood of v naught I can choose to be a open ball of course, I can take it smaller than then I can take the closed ball so no problem. But for m prime for neighborhood of y naught y is some arbitrary space. So, just some number I do not have any balls there yet ok for every y comma v inside m prime coefficient is true. Now, use the continuity of f v naught we can replace m prime by a smaller neighborhood if necessary so that f of y v naught is less than rho by 2 for every y inside m prime. So, there is a choice of m prime at the second stage. First stage is some m prime that is replaced by smaller m prime. This m prime will depend upon the row. Row is chosen in the first instance then I am choosing m prime in a smaller than that. So, choice of m prime will depend upon the row to make it less than rho by 2. See here if I had kept originally t then I had to write a norm of t inverse here ok. So, I have I have this becomes easier because just rho by 2 I can test it all right. Next I come to take a fix take some point m prime and y in m prime ok m prime has been cut down neatly. Now, put s y of v put s of y v remember what was s of y v s of y v by definition is v minus f of y v ok. So, so now I am just writing this s y of v wherein thinking this as a function of v y is fixed s of y v. Now, the derivative of s of y with respect to v is nothing but you see v derivative identity this part is f this is g g of y v prime ok. I am just using this form this definition here it is derivative of this one is identity minus derivative of f all right. So, identity minus g of y v prime. Therefore, the norm of this which is less than equal to norm of identity minus g of this one that is less than equal to c epsilon see g minus identity less than c epsilon that is 14. That is the choice of m prime here you give the first choice second choice is smaller than the whole that is also l ok. So, I have this one less than equal to identity minus c epsilon it less than equal to half because c epsilon is minimum of epsilon and half for every y and v prime belonging to m prime cross the closed ball b rho v 1 ok. Once the derivative less than or to half we know that s y of v 2 minus s y of v 1 is less than equal to this constant lambda 1 by 2 times v 2 minus v 1 ok right. So, this is the mean value theorem that you have proved I mean mean may be inequality for every y v prime belonging to m prime cross b rho bar of you know what is this one this constant is less than 1. Therefore, this is a contraction mapping ok. In order to apply the contraction mapping principle we have yet to show that this s y take the closed ball inside the closed ball. Then the closed ball in a Banach space is a complete metric space on its own then we can apply contraction mapping ok. So, next step we have to do this one. So, here I have made a remark the choice of m prime depends upon rho ok. So, we can see that because the second step here we have already chosen rho and now we are using the continuity of this f v prime the second in the second slot. So, make it as the second the first plot smaller the second slot is fixed here as a function of y which has become smaller that is all ok. So, we have to prove that the closed ball goes inside the closed ball under s y ok. So, that we can think of s y as a contraction mapping inside this metric space which is a complete metric space. So, why this is true take any v such that norm of v minus v naught is less than or equal to rho that means a point of the ball. Then s y of v minus v naught I should say that this also has equal to rho. So, that will prove this statement. So, I am looking at the norm of this one. Now, here you add and subtract s y of v minus s y of v naught minus s y of v naught what is this? This minus minus minus v naught. So, s y of v minus s y of v naught plus this is f of y v naught ok. So, f of y v naught look at this definition f is s y minus this one this is minus of minus that is will plus norm when you take they are the same f of y v naught is this ok we bring that this one this one on this side. So, v minus this one ok what I am saying here s of y v naught here f y v naught minus v will be equal to f of y v naught with a negative sign the norm will be the same ok. So, this norm we have seen is already less than equal to v minus v naught by 2 this norm is less than equal to rho rho by 2. So, that is the second choice here where is the second choice that we have made here f of y v naught norm is less than rho by 2 ok. So, rho by 2 rho by 2 is less than equal to minus some of this one is less than rho ok. It started with v minus v naught is less than equal to rho here. So, norm of v minus v naught by 2 is rho by 2 ok. So, what we have proved is that s y for each y fixed y is a contraction mapping of the closed ball b bar rho v naught ok. So, now we can come to the proof of the statement A that is the step 3. v is a Banach space every closed ball in it is a complete metric space. Therefore, by step 2 we can apply a contraction mapping theorem to conclude that s y has a unique fixed point. We define g by m prime g is m prime 2 m prime is the points where in y varies right into b bar rho by the formula s y of g y is g y ok. So, fixed point of s y for each y there is only one unique map that is important that is one unique point inside this ball. By definition s y of v this is equivalent to saying that f y g y is 0 because s y of v is nothing but v minus f of v s v right. So, f of y g y is 0 will be s of y will be v. So, it follows that for each y in m prime g y is unique for each y number g y is unique that is the conclusion of the contraction mapping in particular g y not has to be v not because y not is already going to v not under f y f of v not y not v not is 0 that was the starting hypothesis. So, g of y not has to be v not the continuity of g is a direct consequence of part b of the contraction mapping theorem which we have proved in the first part in the first module ok. This is s y is continuous in y then this is continuous was following that is what we have proved. So, this proves part a all right proof of part b since now we now assume that y is a Banach space second part we may assume that m prime is a convex neighborhood of of this y not of this y not in both this 14 and 15 right in the beginning you can assume this is a convex in smaller things also you can choose again convex set. So, here in these two ok. So, choice of m prime being convex does not make sense in an arbitrary topological space. So, we could not have said right here once we know that y is a Banach space we can make this one a convex set also ok convex neighborhood of the point y not. So, having made that demand on m prime let us continue now for the proof of this part ok here ok by theorem 1.21 now because of convexity along with this hypothesis 14 what happens f of y v minus f of y v not minus v minus v not see there was a t here now t is identity it is less than it to c epsilon times v minus v not for every y v in m prime cross b bar of rho ok. So, this was the main value in equality ok we have proved this theorem 1.21 y belonging to m prime put v equal to gy in this formula then f of y gy minus f of y v not minus v is gy right gy minus v not is less than or to c epsilon times gy minus v not. Therefore, f of y v not plus gy minus f of y v gy plus v not is there. So, I am taking only this part now f of y v not plus gy minus v not is less than or to c epsilon times gy minus v not why because f of y gy is 0. So, because I am taking a norm I can convert all these negative signs into positive signs ok this minus sign will be this plus sign will be minus sign. So, f of y gy is 0 by choice of gy ok. So, part b of theorem 1.1 namely the continuity part taking f equal to s and v equal to g we get gy minus v not is less than or equal to distance between gy and v not which is distance between gy what is v not is gy not that is less than or to 1 divided by 1 minus epsilon see I am going to use the full statement of part b of theorem 1.1 which I have told you namely the inequality that we have established there norm of gy minus v not if we did not write it in the terms of norm but we have to write it in the matrix notation is gy v not which is gy g v not this was less than or equal to 1 divided by 1 minus epsilon distance between gy and s of y v not. Now go back figure this distance is given by the metric here this metric is given by the norm here. So, I can replace it by norm 1 divided by 1 minus c epsilon gy minus s of y v not ok, but this is by definition f of y v not definition of s ok. Now you use the fact that c epsilon is the minimum of epsilon and 1 by 2. So, it is less than or equal to 1 by 2. So, this 1 divided by 1 minus epsilon is less than or equal to 2 ok. So, this whole thing is less than or equal to twice f of y v not. Therefore, go back now gy minus v not plus f of y v not ok that will be less than twice c epsilon times f of y v not all that we want is some constant here ok depends upon c it may be 3 times may be 4 times that does not matter ok. So, let us write f of y v not as h of y not v not times operating upon y minus v not plus the remainder why we can write it as because this is the derivative of f ok with respect to the y coordinate v not is fixed here ok y minus y not I am taking because it is the derivative with respect to the y coordinate here. So, r y the remainder after n terms after first term this has a property that r y divided by norm of y minus y not tends to 0 as y tends to 0 ok. So, I am using the increment theorem here. Therefore, if you use this in this inequality that we have established what we get is gy minus v not plus h of y not v not minus y minus y not this term ok. So, I have replaced f y this plus this one. So, gy minus v not plus this term is f y minus r y. So, this term I have taken is gy minus v not is as it is this term is replaced by f y minus r y ok, but this part we have already shown twice c epsilon plus f y v not. Therefore, I can add plus norm of r y this is triangle inequality ok this term is as turn equal to this term plus this term. See twice c epsilon f y v not norm plus norm of r y, but that is same thing as now we go back f y v not is this plus this. So, h of y not v not y minus y not plus twice c epsilon plus 1 plus 1 more here this twice c epsilon it comes from here and one here one extra term comes from here. So, this much plus into r y ok dividing out by y minus y not the y minus y not term comes from here norm this is a linear map right. So, norm of this is less than equal to norm of this into norm of y minus v not ok. You divide out by this take the limit what we get this divided by take divided by y minus y not take the limit this is just 0 because this is the derivative of this one ok. So, we get gy minus v not plus h of this y minus v not where y minus y not and y tends to this one ok is less than equal to you see epsilon twice epsilon y minus v not y this is there plus r y is there divided by y minus y not this tends to 0 no I mean does not matter what this constant is right. So, this part vanishes this part will remain ok, but it is twice c epsilon times norm of h y y not v not the norm of y minus y not term cancels out ok. So, I repeat when you take the limit this term vanishes because r y divided by y minus y not tends to 0 this term y minus norm of y minus y not comes out and that has gone down divided that term remains ok. So, this is true for all epsilon positive right that is what I told some c epsilon twice three times whatever you do not care. So, we get that for every epsilon I have taken c epsilon to be minimum of epsilon and half ok. So, if this is true then this left hand side limit must be 0. What is this? This limit tells you that g is differentiable at v not with h as its derivative that is the definition of differentiation sorry minus h as the derivative there is a plus sign here ok. So, this completes the proof of part b and thereby completes the proof of the implicit function theorem. I recall that in the original statement there was a t inverse here, but now in the modify statement we have made t to be identity map that is why the t does not appear here ok. So, that is the proof of implicit function theorem. Now, let us go to inverse function theorem there is one step ahead, but this is the crux of the fact this is the main thing that we want to prove finally v and w are Banach spaces u is an open subset of the first Banach space v f is a function from this open set into w. The condition on f is that it is continuously differentiable function and its derivative at a particular point v not is a similarity ok. The conclusion is that there exists a neighborhood of v not such that f on that neighborhood to its image is a homeomorphism f n is open in w moreover f inverse is also continuously differentiable. So, starting with just a continuously differentiable function which is such that at one point the derivative is invertible in a small neighborhood the function itself is a homeomorphism actually a diffeomorphism because inverse is also continuously differentiable moreover on to the image is also open both n and f n are open n is open in v and f n is open in v. The hypothesis that f df v not is similarity automatically implies that v and w are similar spaces ok. So, how do we prove that since the set of all similarities from one vector space one Banach space to another Banach space is an open subset ok of the continuous linear maps from v to w all continuous maps invertible maps are an open set this is what we have seen. df from u to b v w is given to be continuous ok by replacing u by a smaller neighborhood of v not if necessary we can assume that df v is similar for all v inside u ok all that I have made is you know the map is continuous ok df and at v not it goes to a similarity similarity is contained inside an open subset here we have denoted by a a v w. So, you take an open subset going inside that by continuity and that open subset let us now rewrite it as u we do not want the bigger review at all. So, the assumption is that starting with one at one at one point it is invertible we are assuming that for all v inside u df v is invertible is a similarity ok. Next step in the implicit function theorem above we take y equal to w ok. So, we are in the part b already remember part b of implicit function theorem wanted us to be that y to be a Banach space. So, we are making much special case y as a w itself ok take y not as f of v not v not is given now what is y not do not worry worry about other points take y not f of v not. Now, you take capital F from w to u w cross u to w. In fact, some neighborhood of y not and v not I should take to w, but I can define it this map from the whole of w cross u w given by f of w v is w minus f v very simple function ok capital F this is what we are going to apply the implicit function theorem. Then f is continuously differentiable as a function of v that is what we wanted first of all. In fact, this is continuously differentiable even in terms of w also. So, all the hypothesis that we needed are satisfied for each w inside w the derivative of this function namely when w is fixed at v going to f of w v is minus of df and minus of df at v not is a similarity. So, all the hypothesis of implicit function theorem are satisfied the first part says there is a neighborhood m prime of w of f v not and a row positive such that for each w inside m prime there is a unique g w there are two y y y and so on now y is w. So, I am writing that g w belonging to b row bar such that f of w g w is 0, but what is f of g w w g w it is w minus f of g w w minus f of v right v is g w. So, what is the meaning of this it just means that f g is identity on m prime right. So, that is the meaning of this one moreover the part a already tells you that this map g from m prime to b bar is a unique one and part b says it is continuously differentiable at it is continuous on the whole of sorry it is differentiable at w not. The second part says it is differentiable at w not. The first part says it is continuous ok not continuous differentiable sorry that is not correct. So, that is what we have. So, this is all the implicit function theorem applied to this special case ok. So, are we through? So, we have to see what it is happening here. The existence of g implies this m prime is inside the image of f image of f restricted to the closed ball. See what it means is there exists for each each see map is from m prime f f is a map from v to some neighborhood of right. So, something inside w, but I want to say that m prime is covered by f each point m prime namely w there is a point here which we call the g w f of that is m prime f of that is w. So, this means that m is contained inside the image of f or image of b bar under f ok that is the meaning of like this something. Therefore, what we take is n to be this open ball intersection with f inverse of m prime m prime is an open set open set. So, f inverse of m prime is an open set intersect with the open ball that is an open subset of v. Clearly it is a neighborhood of v not because v not is inside f inverse of f ok f of v not is our starting point this y not or w not whatever ok. So, this is a neighborhood of v not the uniqueness of g implies that this map now restricted to n from n to f n is a bijection because f composite g is identity told you that g is a left inverse of f or f is the right inverse of g ok. But now g is unique ok. So, f must be injective also. So, f is a bijection with g as its inverse ok y not. So, I have already told you that v not is inside n and n is open inside v f from n to f n is a homeomorphism. But why f n is open but f n happens to be just g inverse of m prime because g composite f is identity ok. So, f n is also an open subset of course, it is a neighborhood of f v not ok. So, here is the picture I have drawn v 2 w this f is a multi valid function it is not it is not assumed to be 1 1 map or anything there is no need for that ok. So, what we started we started with a neighborhood m prime and a neighborhood b rho bar here of corresponding neighborhood ok. For each point in m prime there is a unique g inside this bar ok what is g f of that is back to m prime here is w. So, go g and come back by f f of composite g is identity map of this bar which means that this m prime is covered by the image of this one this is some larger thing ok. If you take f of this it could be larger it covers m prime ok. But for points of m prime inside this one there is only one point which is coming to that that is the that is the uniqueness part. If there are another point here coming here the uniqueness will fail right, but some point here may come here some point may here may come here I do not care inside this ball open ball there is only one g ok. So, therefore, when you take this m prime and inverse of that ok inside this one there may be some other point I am intersecting it v in the open ball that is what I am calling it as n on n to m prime f is a bijection now and its inverse is g ok. And what we know is that g is differentiable at v naught here sorry f v naught here that is w naught ok g is continuous f is continuous they are inverse of each other homeomorphism n is open m prime is open. But I am not taking whole of m prime here what I am taking is f of n ok f of n is open alright f of n is also a neighborhood of f v naught. So, only thing that remains is why this g is differentiable on the whole of n we know it only at one point right what is that point that point was an arbitrary point of this neighborhood ok. The hypothesis is true now for all of the points in n remember that was the starting point of our choice of the neighborhood here ok for all we cut down the neighborhood u itself such that d f is invertible on all over of this. So, that hypothesis is there therefore, for every point v prime here I can apply the same theorem to conclude that the inverse exist in some neighborhood the inverse being a unique map ok and it will be differentiable at f of v prime. But inside this neighborhood g has to be the same map because it has to be the inverse of f because f is already one one map therefore, g is differentiable at all the points of f n f n starting with any point here the same hypothesis is applicable here ok. So, this is the last thing I repeat here so far we had only proved that g is differentiable at f v naught. But then the same argument applied at each point w prime is f v prime where v prime range is over n we will tell you that in some smaller neighborhood all that is there in the background we can ignore them. f v prime contained inside f of n everything f is a continuous inverse which is differentiable at f of v prime f has a continuous inverse that but continuous inverse itself is the same g now there is no other because f is already one one map but the inverse of f has to be j on all of f n therefore, g is differentiable at f on the whole of f n ok. So, final thing is that if you take g composite f which is identity its derivative is also identity right identity equal to d of g composite f but all the points you but by chain rule this is d g at f u composite with d f at u which implies if these two functions one composite other is identity d g of f u is nothing but f u d f u inverse ok and that is equal to by definition of our eta eta of d f u ok. Therefore, the continuity of this d g ok follows from the assumption that d f is continuous and the fact that eta is continuous ok as seen in theorem 1.14. Therefore, the derivative is continuous see until here we only show that g is differentiable if d g is continuous follows but d g is given by by this formula namely it is d f inverse. So, I said it is inverse of this one but this is also true for if you take d of f composite g f composite g is already identity. So, it is both ways therefore, one is the inverse of the other ok the inverse taking inverse itself is a continuous operation therefore, we have that d g as a continuous function ok. So, that completes the argument completes all the proofs of all the assertions of the inverse function theorem. So, so theorem 1.14 is proved. So, that is all today. So, let us stop here.